Question

Q C Two identical loudspeakers 10.0 m apart are driven by the same oscillator with a frequency of f = 21.5Hz (Fig. P 18.11) in an area where the speed of sound is 344 m/s. (a) Show that a receiver at point A records a minimum in sound intensity from the two speakers.

Answer 1

The minimum sound intensity can be record at a distance of 10.72m from first speaker

The Distance between two speakers, d =14.9m

Frequency =26.3Hz

Speed of sound V=344.1m/s

The formula to calculate path difference is given as:

d1-d2=λ/2

Where

d1= distance from the first speaker

d2=distance from the second speaker

So we then say that the minimum intensity be at x

Distance of the first speaker from A = X

Distance of the second speaker from A = (14.9-x)m

The wavelength =

V=fλ

λ= V/f

λ= 344.1 / 26.3

λ =13.1m

From the equation above,

d1-d2= λ/2

x-(14.9-x)=13.1/2

x - 14.9 + x =6.54

2x = 6.54+14.9

2x =21.44

x =21.44/2

x = 10.72m

Answer 2

Two identical loudspeakers 10.0m apart driven by the same oscillator with a frequency of 21.5Hz will produce a minimum sound intensity at point A due to destructive interference.

Step-by-step explanation:

To show that a receiver at point A records a minimum sound intensity from two identical loudspeakers, we need to consider the concept of sound interference. When two waves meet, their amplitudes add together, creating regions of constructive interference with higher intensity and regions of destructive interference with lower intensity.

In this case, the two speakers are driven by the same oscillator with a frequency of 21.5Hz. Since the speakers are identical and 10.0m apart, the waves they produce will reach point A with the same amplitude. Therefore, the waves will interfere destructively at point A, resulting in a minimum sound intensity.